The total number of four digit numbers such that each of the first three digits is divisible by the last digit, is equal to ______.

We need to find the total number of four-digit numbers such that each of the first three digits is divisible by the last digit.

Let the four-digit number be $$\overline{abcd}$$ where $$a$$ is the thousands digit, $$b$$ the hundreds, $$c$$ the tens, and $$d$$ the units digit. The conditions are $$d | a$$, $$d | b$$, $$d | c$$, $$a \geq 1$$, and $$d \geq 1$$.

d = 1: Any digit is divisible by 1, so $$a \in \{1,...,9\}$$ (9 choices), $$b, c \in \{0,...,9\}$$ (10 choices each), giving $$9 \times 10 \times 10 = 900$$.

d = 2: $$a \in \{2,4,6,8\}$$ (4 choices), $$b, c \in \{0,2,4,6,8\}$$ (5 choices each), giving $$4 \times 5 \times 5 = 100$$.

d = 3: $$a \in \{3,6,9\}$$ (3 choices), $$b, c \in \{0,3,6,9\}$$ (4 choices each), giving $$3 \times 4 \times 4 = 48$$.

d = 4: $$a \in \{4,8\}$$ (2 choices), $$b, c \in \{0,4,8\}$$ (3 choices each), giving $$2 \times 3 \times 3 = 18$$.

d = 5: $$a \in \{5\}$$ (1 choice), $$b, c \in \{0,5\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 6: $$a \in \{6\}$$ (1 choice), $$b, c \in \{0,6\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 7: $$a \in \{7\}$$ (1 choice), $$b, c \in \{0,7\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 8: $$a \in \{8\}$$ (1 choice), $$b, c \in \{0,8\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

d = 9: $$a \in \{9\}$$ (1 choice), $$b, c \in \{0,9\}$$ (2 choices each), giving $$1 \times 2 \times 2 = 4$$.

$$900 + 100 + 48 + 18 + 4 + 4 + 4 + 4 + 4 = 1086$$ Therefore, the answer is $$\textbf{1086}$$.